User blog:Allam948736/Reptend hierarchy
I invented a hierarchy similar to the fast-growing hierarchy, except the functions have two arguments instead of one, and f0(a, b) is equal to the reptend in the reciprocal of a^b, or ((10^(p*(a^(b-1)))) - 1)/a^b (where p is the period of 1/a). (Except if a is 3, when it is the reptend in the reciprocal of 3^(b+2), because the reptend in 1/3 is 3 and the reptend in 1/9 is 1). If a has no factors that are coprime to 10, then the result is just zero. r0(3, 1) = 37 (1/27 = 0.037037037...) r0(3, 2) = 12345679 (1/81 = 0.01234567901234567901234...) r0(3, 3) = 4115226337448559670781893 r0(3, 4) = 1371742112482853223593964334705075445816186556927297668038408779149519890260631 r0(3, 5) ~ \(4.5724737082761774 \times 10^{239}\) r0(7, 1) = 142857 r0(7, 2) = 20408163265306122448979591836734693877551 r0(7, 3) ~ \(2.91545 \times 10^{291}\) r0(11, 1) = 9 r0(11, 2) = 8264462809173553719 r0(11, 3) ~ \(7.5131 \times 10^{238}\) r1(a, b) is equal to r0 applied b times to b with a as the first argument, just as f1 in the FGH is f0 applied to n n times. r1(3, 1) = r0(3, 1) = 37 r1(3, 2) = r0(3, r0(3, 2)) = r0(3, 12345679) = \(10^{3^{12345679}}\) - 1 / \(3^{12345681}\) = 1552069995.........4587456533 (7.16*10^5890385 digits) r1(3, 3) = r0(3, r0(3, r0(3, 3))) = r0(3, r0(3, 4115226337448559670781893)) r1(7, 1) = r0(7, 1) = 142857 r1(7, 2) = r0(7, r0(7, 2)) = r0(7, 20408163265306122448979591836734693877551) = 3990206661............0856598393 (> \(10^{10^{40}}\) digits) Similarly, r2(a, b) is equal to r1 applied to b b times with a as the first argument, and r3(a, b) is equal to r2 applied b times to b with a as the first argument. r3(3, 3) can actually be shown to be greater than 3^^^^3 (the first term in the sequence defining Graham's number), and ends with ...63929599999999999999999...999999999997 with an unfathomably vast number of 9s. The latter is a number that I named the trireptri. (the initial tri comes from that there are three 3s in the expression, rept is part of reptend, and the second tri comes from the 3s). The trirephept is equal to r7(7, 7). I also invented the generalized reptend function based on this. I defined it like so: r(a) = r0(a, 1) (the reptend in the reciprocal of a unless a is 3, then it is 37) r(a, b) = r(a, b, 0) = r0(a, b) r(a, b, c) = rc(a, b) r(a, b, c, 0) = r(a, b, c) r(a, b, c, d) = r(a, b, r(a, b, c-1, d), d-1) r(a, b, 0, c) = r(a, b) (If one of the arguments is zero, remove all arguments to the right) r(a, b, c, d, e) = r(a, b, c, r(a, b, c, d-1, e), e-1) r(a, b, c, d, e, f) = r(a, b, c, d, r(a, b, c, d, e-1, f), f-1) r(a, b, c, d, e, f, ..., u, v, w, x, y, z) = r(a, b, c, d, e, f, ..., u, v, w, x, r(a, b, c, d, e, f,...,u, v, w, x, y-1, z), z-1) The tetrareptri is r(3, 3, 3, 3), and can actually be shown to be far greater than Graham's number. r(a, b, c, d) has a growth rate of roughly f_w+d© in the fast-growing hierarchy as c increases (r(3, 3, 64, 1) is already greater than Graham's number). I haven't really tested the 5-argument function yet, but its growth rate is probably roughly f_2w+e(d) in the fast-growing hierarchy where e is the fifth argument. The function has a growth rate of roughly f_w^2(n) when the number of arguments is increased (on par with Conway's chained arrow notation), meaning any number of arguments is eventually dominated by f_w^2(n). To go further, I defined the two-comma function, where r(a,,1) is equal to a, and r(a,,b) is r(a, a, a, a, a, ..., a, a, a, a, a) w/ b-1 copies of a. And finally, there's the reptoogol, which is r(r(r(...(r(r(3, 3), 3)...), 3), 3), 3) with 100 3s. Despite the fact that this number is greater than 10^^100, we can still know its first digits, which are 6361454772395...... Category:Blog posts